Let X_{1},\cdots,X_{n} be a random sample from f(x∣\theta)=\thetx^{\theta-1} for 0<x<1.

dooporpplauttssg 2022-04-29 Answered

Let X1,,Xn be a random sample from f(xθ)=θxθ1 for 0<x<1. Find a confidence interval for θ using as pivotal quantity a function of the maximum likelihood estimator for θ.

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Answers (1)

Makayla Santiago
Answered 2022-04-30 Author has 19 answers
Step 1
Assuming you want a confidence interval for θ, you may use the CLT or a known chi distribution for variances and some integrals to get an interval for the parameter. First, we have that
θ^=nln xi.
For the expected value we have the integral
E[X]=01θx(θ1)x dθ=θ1+θ
And for the second moment
μ2=01θxθ1x2 dθ=θ2+θ
So the variance is
Var(X)=θ2+θ(θ1+θ)2=
=θ(θ+1)2(θ+2).
Now then, we know that the following quotient has a chi-squared distribution, so in this case
C=(n1)S2σ2χ{n1}2  and  C=(n1)S2θ(θ+1)2(θ+2)
where S2=1n1(XiX)2. Finally we can work with the random variable C distributed as a chi squared. We want a confidence interval of (1α). Then
P(χ{n1}2(α2)<C<χ{n1}2(1α2))=1α
denoting that χ{n1}2(β) is the β percentile of the χ2 which may be found in reference tables.
From here we work with the interval only substituting the percentiles with 1Dt and 1Dπ, lower and superior.
1Dl<C<1Ds1Dl<(n1)S2θ(θ+1)2(θ+2)<1Ds
Dl>θ(θ+1)2(θ+2)(n1)S2>Ds
Dl((n1)S2)>θ(θ+1)2(θ+2)>Ds((n1)S2)
(Dl((n1)S2))>θ(θ+1)2(θ+2)>(Ds((n1)S2))
(θ^+1)2(θ^+2)(Dl((n1)S2))>θ>(θ^+1)2(θ^+2)(Ds((n1)S2))
where, taking the before definition and derivation, θ^=nln xi and S2=1n1(XiX)2. When we sent the θ to the other side, we simply used its estimate to work around the problem. (However a mathematician should confirm this is valid!) Either way, if you find a better known distribution that may include θ, then try it and tell us if it is simpler!
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